Problem: Find the slope and y-intercept of the line that is ${\text{perpendicular}}$ to $\enspace {y = -\dfrac{1}{3}x + 2}\enspace$ and passes through the point ${(2, 2)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Solution: Lines are considered perpendicular if their slopes are negative reciprocals of each other. The slope of the blue line is ${-\dfrac{1}{3}}$ , and its negative reciprocal is ${3}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = 3x + b}\enspace$ We can plug our point, $(2, 2)$ , into this equation to solve for ${b}$ , the y-intercept. $2 = {3}(2) + {b}$ $2 = 6 + {b}$ $2 - 6 = {b} = -4$ The equation of the perpendicular line is $\enspace {y = 3x - 4}\enspace$. ${m = 3, \enspace b = -4}$